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We in the last video we took the Macloren series of

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representation of e to the x.

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Now let's do it with a couple of other functions and we'll

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see in a few videos it all fits together like a giant puzzle.

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Let's do cosine of x.

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Let's set f of x, f of x is equal to cosine of x.

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What's f prime of x?

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What's the first derivative of cosine of x?

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Well that just equals minus sine of x.

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Minus sine of x.

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What's the second derivative?

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Well that's just minus times derivative of sine of x.

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So the derivative of sine of x is cosine x, it's

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minus cosine of x.

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And what's the third derivative?

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f 3 of x.

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The derivative of cosine x is minus sine of x, we already

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have a minus here so it becomes positive sine of x.

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What's f 4 of x?

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The fourth derivative of x?

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It equals cosine of x again.

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As we keep taking derivatives we'll keep repeating and the

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pattern will go on, right?

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The fifth derivative of x, the fifth derivative of this

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function, the fourth is the same as a function, so the

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fifth is going to be the same as the first derivative.

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cosine of x is sine of x.

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Hopefully you see the pattern.

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We're going to do the Macloren representation, which is a

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specific example of the Taylor series where we figure out the

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values of the derivatives at x is equal to zero.

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So let's do that right now.

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So f of zero, let me do it in to another color

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to fend off monotony.

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f of zero.

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What's cosine of zero?

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Cosine of zero is 1.

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f prime of zero is equal to sine of-- well not minus sine

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of zero, but what sine of zero?

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Sine of zero is zero, so minus zero is still

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zero, so this is zero.

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f prime prime of zero.

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Cosine of zero we already know is one.

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We have a negative sine here, so it becomes a minus one.

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The third derivative at x is equal to zero.

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Sine of zero is zero.

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So this is zero.

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I think you might start to see a pattern emerging.

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The fourth derivative at zero.

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Cosine of zero is equal to 1.

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And then the fifth derivative, this is hard to read but you

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get the point is just zero again.

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So what's the pattern as we take the derivatives?

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1, zero, minus 1, zero, 1, zero.

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So it alternates between zero and 1.

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So 1, zero, minus 1, zero, positive, zero, negative,

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zero, positive.

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So every other number is zero and in between them we

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alternate between a positive 1 and a negative 1.

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So now let's use that information to figure

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out them the Macloren series representation.

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So we proved, hopefully, we didn't prove it definitely

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converges over the entire domain of the function.

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You have to take my word for it.

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We'll experiment a little bit with a graphing calculator

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in a few videos.

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We said that this representation-- and it should

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make intuitive sense, because when you take the infinite

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Macloren series, when you take that infinite sum, you're

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essentially creating a function where that function is equal to

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your original function at the point you chose.

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In the case of a Macloren we're picking x equals zero, and it

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equals every derivative of this function.

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Just intuitively it seems, well if a function equals something

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at a point and every one of its derivatives is also equal to

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the function at that point, maybe those functions are

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equal to each other.

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I haven't proven that to you yet.

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We know that the representation is a sum from n is equal to

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zero to infinity of the nth derivative evaluated at zero.

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A Macloren series is a specific case of a Taylor series.

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We actually haven't done anything with Taylor series, I

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was hoping to get there later.

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But the Macloren series is a really cool one because it's

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going to show us all these relationships between e and

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cosine and sine and eventually i and pi and you will

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find it exciting.

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The Macloren is that times x to the n over n factorial.

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That's what we said it was.

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So if this is our f of x, f of x is cosine of x, what

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does this turn into?

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Well, f of x is equal to, it equals f of zero times x to

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the zero over zero factorial, that's just one, right?

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Plus, now we're at n equals 1.

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It's the first derivative at zero.

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f prime of zero, well that's just equal to zero.

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And who cares what that-- that would be x to the

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first over 1, right?

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Now we're at the second derivative.

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The second derivative at zero is minus 1.

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Minus 1 times x squared over 2 factorial plus the

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third derivative at zero.

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The third derivative at zero we figured out was zero.

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Zero who cares what that is.

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It would have been x to the third over 3 factorial.

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And then what's the fourth derivative?

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The fourth derivative at zero is just equal to 1.

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So we have times 1 and then we're at x to the fourth

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over 4 factorial.

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Let me see if I can write this a little bit neater.

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The next one, the fifth derivative at zero times x to

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the fifth over 5 factorial.

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We'll keep going.

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Let me write this, clean this up and hopefully the pattern

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merges if it hasn't emerged already.

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f of x is equal to cosine of x is equal to-- let met get rid

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of the zeros-- 1 and then we have minus x squared

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over 2 factorial.

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This term, this goes away.

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This is a zero term.

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And the next one is a positive.

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Plus x to the fourth over 4 factorial.

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And the fifth term goes away.

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But then the cycle continues.

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The next one is going to be minus.

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Because we had minus 1 plus 1.

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It's going to be minus x to the sixth over 6 factorial.

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You could take the sixth derivative.

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You'll see that the derivative of minus sine of x is minus

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cosine of x, that's where we get the minus 1 from.

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And they we're going to go plus.

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So we're just taking all the even terms. x to the eighth

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over 8 factorial minus x to the 10th over 10 factorial.

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We could just keep going on and on and on.

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And so we have a situation where we can rewrite cosine of

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x is equal to the sum, if you believe that this Macloren

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series actually does converge to cosine of x over the entire

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domain of x, that's kind of an assumption we're making.

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Hopefully one day we will have the tools set to actually

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prove that as well.

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From n is equal to zero.

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So what's happening?

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We're taking all of the even powers.

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So we could say x to the 2n, that ensures that no matter

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what value of n I put in here I get an even numbers.

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So we'll go to the zeroth power then the second

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power, over 2n factorial.

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So that takes care of going from 1 to x squared over 2,

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to x to the fourth over 4 factorial, 6 over 6

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factorial, et cetera.

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But now we have to make it switched signs like that.

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Well let's just multiply it negative 1.

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Let's see what we can do.

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Negative 1 to the-- so we want the first term to be positive,

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the second term to be negative.

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So we could say times minus 1 to the n plus 1.

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Let's see if that works.

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When m is zero what's negative 1 to the n plus 1?

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zero, it would be minus 1.

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And then when it's 1-- When it's zero-- no, no it's just

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going to be negative 1 to the n.

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Because when it's zero, negative 1 to zero is 1.

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When it's 1, negative 1.

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So this will work out.

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Negative 1 to the n is cosine.

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You could try it out.

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This is the n is equal to zero.

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We need to switch colors.

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That's n is equal to zero and here we get x to the zero over

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zero factorial, which is 1.

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We have negative 1 to the zero is 1, so that becomes 1.

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When n is equal to 1, this becomes x squared over 2

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factorial, we have negative 1 to the 1 power, so that's

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where you get the negative 1.

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And then when n is equal to 2, the negative 1 squared

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becomes positive again.

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So the negative 1 is what provides the

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alternating numbers.

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So pretty neat.

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We just figured out another way to represent cosine of x.

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And it might be looking a little bit interesting to you

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that this representation kind of resembles part of the

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representation of e to the x.

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What's the difference between this and the e to the x? e to

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the x had the odd exponent terms and it didn't

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switch signs.

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But other than that, they're pretty much the same.

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So in the next video we'll do sine of x and then we'll

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try to put it all together.

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I'll see you soon.

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